SEMISIMPLICITY KEITH CONRAD A subspace W of an F -vector space V always has a complementary subspace: V = W ⊕W 0 for some subspace W 0. This can be seen using bases: extend a basis of W to a basis of V and let W 0 be the span of the part of the basis of V not originally in W . Of course there are many ways to build a complementary subspace, since extending a basis is a rather flexible procedure. If the vector space or subspace has extra structure then we can ask if a complement to W can be found with properties related to this structure. For example, when V = Rn we have the concept of orthogonality in Rn, and any subspace W has an orthogonal complement: Rn = W ⊕ W 0 where W ? W 0, and moreover there is only one such complement to W . The orthogonal complement is tied up with the geometry of Rn. Another kind of structure we can put on subspaces (of general vector spaces) is stability under a linear operator on the whole space. Given a linear operator A: V ! V , a subspace W satisfying A(W ) ⊂ W is called an A-stable subspace. For example, a one-dimensional A-stable subspace is the same thing as the line spanned by an eigenvector for A: if W = F v is A-stable then A(v) = λv for some λ 2 F , so v is an eigenvector. We ask: does an A-stable subspace have a complement which is also A-stable? Example 1. If A = idV then all subspaces are A-stable, so any complement to an A-stable subspace is also A-stable. In particular, an A-stable complement to a subspace is not unique (if the subspace isn't f0g or V ). 1 1 2 x Example 2. Consider A = ( 0 1 ) acting on F and its eigenspace W = f 0 : x 2 F g. This is A-stable. A complementary subspace to W would be 1-dimensional and thus also be spanned by an eigenvector for A, but W is the only eigenspace of A. So W is A-stable but has no A-stable complement. Remember this example! From now on, all linear operators are acting on nonzero finite-dimensional vector spaces. While a subspace stable for an operator does not always have a stable complementary subspace, we will show any stable subspace has a stable complementary subspace when the operator is potentially diagonalizable. We will carry out the proof in the diagonalizable case first since the ideas are a simpler there, and then one appreciates more clearly the extra details that crop up in the more general potentially diagonalizable case. Lr Theorem 3. Let A: V ! V be diagonalizable and V = i=1 Eλi be the corresponding eigenspace decomposition. Lr (1) If W is an A-stable subspace of V then W = i=1(W \ Eλi ) and each W \ Eλi is A-stable, (2) Any A-stable subspace of V has an A-stable complement. Pr Proof. (1) We will show W = i=1(W \ Eλi ); the sum is automatically direct since the subspaces Eλi 's are linearly independent. (Eigenvectors for different eigenvalues are linearly independent.) For w 2 W , write w = w1 +···+wr with wi 2 Eλi . We will show the eigencomponents wi Pr all lie in W , so wi 2 W \ Eλi for all i and thus W = i=1(W \ Eλi ). The reason wi 2 W is 1 2 KEITH CONRAD that W is h(A)-stable for any h(T ) 2 F [T ] since W is A-stable, and in the proof that V has an eigenspace decomposition for A it is shown that wi = hi(A)(w) for a certain polynomials hi(T ) 2 F [T ]. Since W and Eλi are both A-stable, so is their intersection W \ Eλi . Lr (2) Let W be A-stable and Wi = W \ Eλi , so W = i=1 Wi by (1). In each Eλi , A acts by scaling by λi, so all subspaces of Eλi are A-stable. (Not all subspaces of the 0 whole space V are A-stable!) Let Wi be any subspace complement to Wi inside Eλi . Then 0 Pr 0 Lr 0 0 W := i=1 Wi = i=1 Wi is a subspace of V that is A-stable (because each Wi is A-stable) and r r 0 M 0 M W ⊕ W = (Wi ⊕ Wi ) = Eλi = V: i=1 i=1 Although a potentially diagonalizable operator A: V ! V may not have eigenspaces in V , its minimal polynomial has distinct irreducible factors and we can use them to extend the previous theorem to the potentially diagonalizable case. Theorem 4. Let A: V ! V be potentially diagonalizable, with minimal polynomial mA(T ). For each monic irreducible factor πi(T ) of mA(T ), let Vi = fv 2 V : πi(A)(v) = 0g. Lr (1) Each Vi is A-stable and V = i=1 Vi. Lr (2) If W is an A-stable subspace of V then W = i=1(W \ Vi) and each W \ Vi is A-stable. (3) Any A-stable subspace of V has an A-stable complement. If A is diagonalizable, so each πi(T ) is linear, say πi(T ) = T − λi, then Vi = Eλi is an eigenspace and this theorem becomes Theorem 3. Proof. (1) Since A and πi(A) commute, if v 2 Vi then A(v) 2 Vi. Therefore A(Vi) ⊂ Vi for all i, so each Vi is A-stable. We will show that it is possible to \project" from V to Vi using a polynomial in the operator A. We seek h1(T ); : : : ; hr(T ) in F [T ] such that (1) 1 = h1(T ) + ··· + hr(T ); hi(T ) ≡ 0 mod mA(T )/πi(T ): Once these polynomials are found, πi(T )hi(T ) is divisible by mA(T ) for all i, so πi(A)hi(A) = O. Then replacing T with the operator A in (1) and applying all operators to any v 2 V gives v = h1(A)(v) + ··· + hr(A)(v); πi(A)hi(A)(v) = 0: The second equation tells us hi(A)(v) 2 Vi, so the first equation shows V = V1 + ··· + Vr. To show this sum is direct, suppose (2) v1 + ··· + vr = 0 with vi 2 Vi. We want to show each vi is 0. Apply hi(A) to both sides of (2). Since hi(T ) is divisible by πj(T ) for j 6= i, hi(A)(vj) = 0 for j 6= i (look at the definition of Vj). Therefore hi(A)(vi) = 0. Also πi(A)(vi) = 0 by the definition of Vi, so hj(A)(vi) = 0 for j 6= i. Thus P idV = hi(A) + j6=i hj(A) kills vi, so vi = 0. It remains to find polynomials hi(T ) fitting (1). Set fi(T ) = mA(T )/πi(T ). These polynomials are relatively prime as an r-tuple, so some F [T ]-linear combination of them is 1: 1 = g1(T )f1(T ) + ··· + gr(T )fr(T ): Use hi(T ) = gi(T )fi(T ). SEMISIMPLICITY 3 Pr (2) We will show W = i=1(W \ Vi). Then the sum must be direct because the Vi's are linearly independent by (1). For w 2 W , the proof of (1) shows that the component of w in Vi is wi := hi(A)(w) for some polynomial hi(T ). Since W is A-stable and hi(A) is a polynomial in A, wi 2 W . Therefore wi 2 W \ Vi. Since W and Vi are each carried into themselves by A, so is W \ Vi. (3) This will be more technical than the proof of the corresponding case for diagonalizable operators. Lr Let W be A-stable and set Wi := W \ Vi, so W = i=1 Wi and the Wi's are A-stable by (2). To find an A-stable complement to W in V it suffices (in fact, it is equivalent) to find an A-stable complement to Wi in Vi for all i. Then the sum of these complements will be an A-stable complement to W in V . Unlike in the proof of Theorem 3(2), A need not be a scaling operator on Vi, so a random subspace complement to Wi in Vi is unlikely to be A-stable. We have to think more carefully to find an A-stable complement of Wi in Vi. Think about Vi as an F [T ]-module where any f(T ) 2 F [T ] acts on V by f(T )(v) := f(A)(v). Since A(Wi) ⊂ Wi, Wi is an F [T ]-submodule of Vi. More generally, the F [T ]- submodules of Vi are precisely the A-stable F -vector spaces in V . We seek an F [T ]- 0 0 submodule Wi of Vi such that Vi = Wi ⊕Wi . Since πi(T ) kills Vi, Vi is an F [T ]=(πi)-module and Wi is an F [T ]=(πi)-submodule. Now F [T ]=(πi) is a field, so Vi is a vector space over 0 F [T ]=(πi) and Wi is a subspace over this field. Set Wi to be any complementary subspace to Wi inside Vi as F [T ]=(πi)-vector spaces. (When det πi > 1, this is a stronger condition 0 than being a complementary subspace in Vi as F -vector spaces.) Since Wi is an F [T ]=(πi)- 0 Pr 0 submodule of Vi, it is an F -vector space and A-stable, so we are done: W = i=1 Wi is an A-stable complement to W in V .